HI6007 Statistics Assignment Sample
Here’s the best sample of HI6007 Statistics Assignment, written by the expert.
Question 1
| Source of variation | Sum of squares | Degrees of freedom | Mean square | F |
| Between treatments | 90 | 3 (k-1) | 30 | 5 |
| Within treatments (Error) | 120 | 20 (N-k) | 6 | |
| Total | 210 | 23 (N-1) | ||
Calculation:
Mean square between treatments = sum of squares / degrees of freedom
= 90/3
= 30
Mean square within treatments = sum of squares / degrees of freedom
= 120/20
= 6
F= Mean square between treatments / Mean square within treatments
= 30/6
= 5
Total for sum of squares= sum of squares between treatments + sum of squares within treatments
= 90+120
= 210
Total of degrees = N-1
k-1= 3
k = 3+1 = 4
N-k= 20
N-4= 20
N= 24
Then, N-1 = 24-1 = 23
Use α = .01 to determine if there is any significant difference among the means.
F(3,20) = 4.94 (at 0.01 significance level)
Fcalculated > Fcritical
We can reject null hypothesis as there is significant difference among the means.
- How many groups have there been in this problem?
k= 4
- What has been the total number of observations?
N= 24
Question 4
| Source of variation | Sum of squares | Degrees of freedom | Mean square | F |
| Between treatments | 3200 | 4 | 800 | 0.9729 |
| Within treatments (Error) | 7400 | 9 | 822.22 | |
| Total | 10600 | 13 | ||
k-1 = 5-1= 4
N=14
N-k = 14-5= 9
Mean square between treatments = sum of squares / degrees of freedom
800= x/4
X= 3200
Mean square within treatments = sum of squares / degrees of freedom
y= 10600-3200/9
y= 7400/9
= 822.22
F= Mean square between treatments / Mean square within treatments
=800/822.22
= 0.9729
Question 5
| Store 1 | Store 2 | Store 3 |
| 46 | 34 | 33 |
| 47 | 36 | 31 |
| 45 | 35 | 35 |
| 42 | 39 | |
| 45 |
| Anova: Single Factor |
|
|||||
| SUMMARY | ||||||
| Groups | Count | Sum | Average | Variance | ||
| Store 1 | 5 | 225 | 45 | 3.5 | ||
| Store 2 | 4 | 144 | 36 | 4.666667 | ||
| Store 3 | 3 | 99 | 33 | 4 | ||
| ANOVA | ||||||
| Source of Variation | SS | df | MS | F | P-value | F crit |
| Between Groups | 324 | 2 | 162 | 40.5 | 3.16E-05 | 4.256495 |
| Within Groups | 36 | 9 | 4 | |||
| Total | 360 | 11 |
The above p value is greater than 0.05 or Fcritical < Fcalculated means there is a significant difference in the average sales of the three stores at 5% significance level (Reject null hypothesis).
Question 6
| Store 1 | Store 2 | Store 3 | Store 4 | Store 5 | |
| Box 1 | 210 | 230 | 190 | 180 | 190 |
| Box 2 | 195 | 170 | 200 | 190 | 193 |
| Box 3 | 295 | 275 | 290 | 275 | 265 |
- Hypotheses
H0: µ1=µ2=µ3
Ha: At least one mean is different
b)
| Anova: Single Factor | ||||||
| SUMMARY | ||||||
| Groups | Count | Sum | Average | Variance | ||
| Column 1 | 5 | 1000 | 200 | 400 | ||
| Column 2 | 5 | 948 | 189.6 | 133.3 | ||
| Column 3 | 5 | 1400 | 280 | 150 | ||
| ANOVA | ||||||
| Source of Variation | SS | df | MS | F | P-value | F crit |
| Between Groups | 24467.2 | 2 | 12233.6 | 53.71111 | 1.03E-06 | 6.926608 |
| Within Groups | 2733.2 | 12 | 227.7667 | |||
| Total | 27200.4 | 14 |
- c) Conclusion
In this, the null hypothesis is rejected as fcritical< Fcalculated means at least one mean is different from the others (Reject null hypothesis). Box 3 will generate the maximum sales as its mean is the largest as compared to others.
Question 7
| Brand A | Brand B | Brand C | |
| Average Mileage | 37 | 38 | 33 |
| Sample variance | 3 | 4 | 2 |
| Anova: Single Factor | ||||||
| SUMMARY | ||||||
| Groups | Count | Sum | Average | Variance | ||
| Column 1 | 2 | 40 | 20 | 578 | ||
| Column 2 | 2 | 42 | 21 | 578 | ||
| Column 3 | 2 | 35 | 17.5 | 480.5 | ||
| ANOVA | ||||||
| Source of Variation | SS | df | MS | F | P-value | F crit |
| Between Groups | 13 | 2 | 6.5 | 0.011916 | 0.988202 | 9.552094 |
| Within Groups | 1636.5 | 3 | 545.5 | |||
| Total | 1649.5 | 5 |
If the p-value is greater than 0.05 or Fcritical>Fcalculated, we cannot reject the null hypothesis that there’s no difference between the means and conclude that a significant difference does not exist (non significant result). It means the mean mileage for all three brands of tires is the same at 5% significance level.
References
Jaggia, S., Kelly, A., Beg, A.B.M., Leighton, C., Olaru, D., Salzman, S. and Sriananthakumar, S., 2016. Essentials of business statistics: communicating with numbers. McGraw-Hill Education.
Todua, N. and Dotchviri, T., 2015. Anova in Marketing Research of Consumer Behavior of Different Categories in Georgian Market. Annals of the “Constantin Brâncusi” University of Târgu Jiu, Economy Series, (1), pp.183-190.
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